Hello : the tangent has same slope of the line 16x -y +5 = 0 because parallel. y = 16x + 5 the slope is : 16 but : f'(x) = 16 f'(x) = 16x^3 - 16x +16 solve : 16x^3 - 16x +16 = 16 x^3 - x +1 =1 x^3 - x = 0 x(x² - 1 ) =0 x(x -1 )(x +1) =0 x=0 or x-1 =0 or x +1= 0 x=0 or x = 1 or x = -1 if : x = 0 : y = 4(0)^4 -8(0)^2+16(0) +7 =7 if : x =1 : y = 4(1)^4 -8(1)^2+16(1) +7 = 19 if : x = -1 : y = 4(-1)^4 -8(-1)^2+16(-1) +7 = -13 <span>the points : A(0,7) , B(1 , 19) , c ( -1 , -13)</span>